I have a model following the PDF $$f_\theta(x) = \theta^2 x \exp(-\theta x)\delta_{[0, +\infty)}(x)$$ where $\theta > 0$ is the parameter and $\delta_\mathcal{S}$ denotes the indicator function of the set $\mathcal{S}$.

I am ask to prove that this model is regular, for which I have to prove that it is dominated. How should I do that?

Below is the definition of "dominated" that I found the the book Theory of Point Estimation by E.L. Lehmann and George Casella:

Statistical problems are concerned not with single probability distributions but with families of such distributions $$\mathcal{P} = \{P_\theta, \theta\in\Omega\}$$ defined over a common measurable space $(\mathcal{X}, \mathcal{A})$. When all the distributions of $\mathcal{P}$ are absolutely continuous with respect to a common measure $\mu$, as will usually be the case, the family $\mathcal{P}$ is said to be dominated (by $\mu$).

Most of the examples with which we shall deal belong to one or the other of the following two cases.

(i) The discrete case. Here, $\mathcal{X}$ is a countable set, $\mathcal{A}$ is the class of subsets of $\mathcal{X}$, and the distributions of $\mathcal{P}$ are dominated by counting measure.

(ii) The absolutely continuous case. Here, $\mathcal{X}$ is a Borel subset of a Euclidean space, $\mathcal{A}$ is the class of Borel subsets of $\mathcal{X}$, and the distributions of $\mathcal{P}$ are dominated by Lebesgue measure over $(\mathcal{X}, \mathcal{A})$.

Thank you in advance for any help!

  • $\begingroup$ can you define what "regular" and "dominated" mean? $\endgroup$
    – jld
    Jan 22, 2018 at 15:31
  • $\begingroup$ @Chaconne I have updated with a definition of "dominated". Thanks. $\endgroup$
    – f10w
    Jan 22, 2018 at 15:39
  • $\begingroup$ thanks for the update. So the definition discusses measures but you've already got a density which mean that, by definition, each of your $f_\theta$ is equal to $\frac{\text d P_\theta}{\text d\mu_\theta}$ for some $\mu_\theta$. This means for each $P_\theta$ you have a dominating measure so you then need to show that all these $\mu_\theta$ are in fact the same measure $\mu$, and that measure dominates $\{P_\theta\}$. Do you know what the $\mu_\theta$ are? $\endgroup$
    – jld
    Jan 22, 2018 at 15:44
  • $\begingroup$ @Chaconne No all I'm given is the above PDF :( Another definition of "dominated": www2.cirano.qc.ca/~dufourj/Web_Site/ResE/… I guess they are equivalent. $\endgroup$
    – f10w
    Jan 22, 2018 at 15:49
  • $\begingroup$ The quotation demonstrates that "dominated by Lebesgue measure" and "absolutely continuous" are synonymous. $\endgroup$
    – whuber
    Jan 22, 2018 at 15:53